Evaluate the following expression. Your answer must be exact. $(\sqrt{2}+\sqrt{2}i)^6=$
The Strategy The easiest way to find $z^{n}$ for a complex number $z=({a}+{b}i)$ is using its modulus and argument. Therefore, our solution will consist of the following steps: Find the modulus and argument of $z$. [How is this done, in general?] Find the modulus and argument of $z^{n}$. [How is this done, in general?] Find the rectangular form $z^{n}$. Find the modulus and argument of $(\sqrt{2}+\sqrt{2}i)$ $({\sqrt{2}}+{\sqrt{2}}i)$ is of the form $({a}+{b}i)$, where ${a=\sqrt{2}}$ and ${b=\sqrt{2}}$. Therefore: $\begin{aligned}r&=\sqrt{{a}^2 + {b}^2} \\\\&=\sqrt{({\sqrt{2}})^2 + ({\sqrt{2}})^2} \\\\&=\sqrt{{2}+{2}} \\\\&=2\end{aligned}$ Using the arctangent formula, we have: $\begin{aligned}\theta&=\arctan\left(\dfrac{{b}}{{a}}\right) \\\\&=\arctan\left(\dfrac{{\sqrt{2}}}{{\sqrt{2}}}\right) \\\\&=45^\circ\end{aligned}$ Since both ${a=\sqrt{2}}$ and ${b=\sqrt{2}}$ are positive, $(\sqrt{2}+\sqrt{2}i)$ lies in Quadrant $1$. Therefore, $\theta$ must be between $0^\circ$ and $90^\circ$, so our answer matches our requirements. Find the modulus and argument of $(\sqrt{2}+\sqrt{2}i)^6$ We found that the modulus and argument of $({\sqrt{2}}+{\sqrt{2}}i)$ are $2$ and $45^\circ$. Therefore, the modulus and argument of $({\sqrt{2}}+{\sqrt{2}}i)^6$ are $2^6=64$ and $(45^\circ)\cdot6=270^\circ$. Find the rectangular form of $(\sqrt{2}+\sqrt{2}i)^6$ Since the argument is $270°$, we know the number lies on the negative side of the imaginary number axis and is therefore a negative pure imaginary number. Since the modulus is $64$, our solution is $-64i$. [What does this look like graphically?] [How do we find this algebraically?] Summary $(\sqrt{2}+\sqrt{2}i)^6=-64i$